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प्रश्न
Find the 25th term of the AP `5, 4 1/2, 4, 3 1/2, 3,`....
योग
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उत्तर
The given AP is `5, 4 1/2, 4, 3 1/2, 3,`....
First term = 5
Common difference = `4 1/2 - 5`
⇒ `9/2 - 5`
⇒ `(9 - 10)/2 = -1/2`
∴ a = 5 and d = `-1/2`
Now, T25 = a + (25 – 1)d
= a + 24d
= `5 + 24 xx (-1/2)`
= 5 – 12
= –7
∴ 25th term = –7
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